Let’s go through each step of the titration for the weak acid (hydrofluoric acid, HF) and strong base (KOH). We need to calculate the pH at different points during the titration.<\/p>\n\n\n\n
Given Data:<\/h3>\n\n\n\n\n- Volume of HF = 56.9 mL = 0.0569 L<\/li>\n\n\n\n
- Concentration of HF = 0.301 M<\/li>\n\n\n\n
- KaK_aKa\u200b of HF = 7.2\u00d710\u221247.2 \\times 10^{-4}7.2\u00d710\u22124<\/li>\n\n\n\n
- Concentration of KOH = 0.301 M<\/li>\n\n\n\n
- Volume of KOH added will vary at different points.<\/li>\n<\/ul>\n\n\n\n
\n\n\n\n1. Before the addition of any KOH (Initial pH of HF)<\/strong><\/h3>\n\n\n\nFor the initial pH of a weak acid, we need to use the expression for the dissociation of HF: HF\u21ccH++F\u2212HF \\rightleftharpoons H^+ + F^-HF\u21ccH++F\u2212<\/p>\n\n\n\n
The equilibrium expression for the dissociation is: Ka=[H+][F\u2212][HF]K_a = \\frac{[H^+][F^-]}{[HF]}Ka\u200b=[HF][H+][F\u2212]\u200b<\/p>\n\n\n\n
Let\u2019s assume the concentration of [H+][H^+][H+] is xxx, and at equilibrium: Ka=x2[HF]\u2212xK_a = \\frac{x^2}{[HF] – x}Ka\u200b=[HF]\u2212xx2\u200b<\/p>\n\n\n\n
For weak acids, xxx will be very small compared to the initial concentration, so we can approximate [HF]\u2212x\u2248[HF][HF] – x \\approx [HF][HF]\u2212x\u2248[HF].<\/p>\n\n\n\n
Thus: Ka=x2[HF]K_a = \\frac{x^2}{[HF]}Ka\u200b=[HF]x2\u200b<\/p>\n\n\n\n
Solving for xxx, the concentration of H+H^+H+: x=Ka\u00d7[HF]x = \\sqrt{K_a \\times [HF]}x=Ka\u200b\u00d7[HF]\u200b<\/p>\n\n\n\n
Substitute the values: x=(7.2\u00d710\u22124)\u00d70.301x = \\sqrt{(7.2 \\times 10^{-4}) \\times 0.301}x=(7.2\u00d710\u22124)\u00d70.301\u200b x\u22481.48\u00d710\u22122\u2009Mx \\approx 1.48 \\times 10^{-2} \\, \\text{M}x\u22481.48\u00d710\u22122M<\/p>\n\n\n\n
Now, calculate the pH: pH=\u2212log\u2061([H+])=\u2212log\u2061(1.48\u00d710\u22122)\u22481.83\\text{pH} = -\\log([H^+]) = -\\log(1.48 \\times 10^{-2}) \\approx 1.83pH=\u2212log([H+])=\u2212log(1.48\u00d710\u22122)\u22481.83<\/p>\n\n\n\n
\n\n\n\n2. After the addition of 14.0 mL of KOH<\/strong><\/h3>\n\n\n\nTo calculate the pH after adding 14.0 mL of KOH, we first determine the moles of both HF and KOH.<\/p>\n\n\n\n
\n- Moles of HF:<\/li>\n<\/ul>\n\n\n\n
moles of HF=0.301\u2009M\u00d70.0569\u2009L=0.0171\u2009mol\\text{moles of HF} = 0.301 \\, \\text{M} \\times 0.0569 \\, \\text{L} = 0.0171 \\, \\text{mol}moles of HF=0.301M\u00d70.0569L=0.0171mol<\/p>\n\n\n\n
\n- Moles of KOH:<\/li>\n<\/ul>\n\n\n\n
moles of KOH=0.301\u2009M\u00d70.0140\u2009L=0.00421\u2009mol\\text{moles of KOH} = 0.301 \\, \\text{M} \\times 0.0140 \\, \\text{L} = 0.00421 \\, \\text{mol}moles of KOH=0.301M\u00d70.0140L=0.00421mol<\/p>\n\n\n\n
KOH is a strong base, and it will neutralize HF, forming F\u2212F^-F\u2212 and H2OH_2OH2\u200bO.<\/p>\n\n\n\n
\n- Remaining moles of HF:<\/li>\n<\/ul>\n\n\n\n
0.0171\u2009mol\u22120.00421\u2009mol=0.0129\u2009mol0.0171 \\, \\text{mol} – 0.00421 \\, \\text{mol} = 0.0129 \\, \\text{mol}0.0171mol\u22120.00421mol=0.0129mol<\/p>\n\n\n\n
\n- Moles of F\u2212F^-F\u2212 formed:<\/li>\n<\/ul>\n\n\n\n
0.00421\u2009mol0.00421 \\, \\text{mol}0.00421mol<\/p>\n\n\n\n
Now, calculate the new concentrations in the total volume of the solution, which is: Vtotal=56.9\u2009mL+14.0\u2009mL=70.9\u2009mL=0.0709\u2009LV_{\\text{total}} = 56.9 \\, \\text{mL} + 14.0 \\, \\text{mL} = 70.9 \\, \\text{mL} = 0.0709 \\, \\text{L}Vtotal\u200b=56.9mL+14.0mL=70.9mL=0.0709L<\/p>\n\n\n\n
\n- Concentration of HF:<\/li>\n<\/ul>\n\n\n\n
[HF]=0.0129\u2009mol0.0709\u2009L=0.182\u2009M[HF] = \\frac{0.0129 \\, \\text{mol}}{0.0709 \\, \\text{L}} = 0.182 \\, \\text{M}[HF]=0.0709L0.0129mol\u200b=0.182M<\/p>\n\n\n\n
\n- Concentration of F\u2212F^-F\u2212:<\/li>\n<\/ul>\n\n\n\n
[F\u2212]=0.00421\u2009mol0.0709\u2009L=0.0594\u2009M[F^-] = \\frac{0.00421 \\, \\text{mol}}{0.0709 \\, \\text{L}} = 0.0594 \\, \\text{M}[F\u2212]=0.0709L0.00421mol\u200b=0.0594M<\/p>\n\n\n\n
At this point, we use the Henderson-Hasselbalch equation to find the pH: pH=pKa+log\u2061([F\u2212][HF])\\text{pH} = \\text{pKa} + \\log \\left( \\frac{[F^-]}{[HF]} \\right)pH=pKa+log([HF][F\u2212]\u200b)<\/p>\n\n\n\n
First, calculate pKa\\text{pKa}pKa: pKa=\u2212log\u2061(Ka)=\u2212log\u2061(7.2\u00d710\u22124)\u22483.14\\text{pKa} = -\\log(K_a) = -\\log(7.2 \\times 10^{-4}) \\approx 3.14pKa=\u2212log(Ka\u200b)=\u2212log(7.2\u00d710\u22124)\u22483.14<\/p>\n\n\n\n
Then, calculate the pH: pH=3.14+log\u2061(0.05940.182)\u22483.14+log\u2061(0.326)\u22483.14\u22120.486\u22482.65\\text{pH} = 3.14 + \\log \\left( \\frac{0.0594}{0.182} \\right) \\approx 3.14 + \\log(0.326) \\approx 3.14 – 0.486 \\approx 2.65pH=3.14+log(0.1820.0594\u200b)\u22483.14+log(0.326)\u22483.14\u22120.486\u22482.65<\/p>\n\n\n\n
\n\n\n\n3. At the half-equivalence point (the titration midpoint)<\/strong><\/h3>\n\n\n\nAt the half-equivalence point, the moles of KOH added will be equal to half the moles of HF.<\/p>\n\n\n\n
\n- Moles of KOH at half-equivalence:<\/li>\n<\/ul>\n\n\n\n
0.01712=0.00855\u2009mol\\frac{0.0171}{2} = 0.00855 \\, \\text{mol}20.0171\u200b=0.00855mol<\/p>\n\n\n\n
The concentration of F\u2212F^-F\u2212 and HFHFHF will be the same at this point, so: pH=pKa=3.14\\text{pH} = \\text{pKa} = 3.14pH=pKa=3.14<\/p>\n\n\n\n
\n\n\n\n4. At the equivalence point<\/strong><\/h3>\n\n\n\nAt the equivalence point, all HF has reacted with KOH. The solution now contains only the conjugate base F\u2212F^-F\u2212, which will hydrolyze in water to form OH\u2212OH^-OH\u2212. The hydrolysis reaction is: F\u2212+H2O\u21ccHF+OH\u2212F^- + H_2O \\rightleftharpoons HF + OH^-F\u2212+H2\u200bO\u21ccHF+OH\u2212<\/p>\n\n\n\n
We can use the Kb of F\u2212F^-F\u2212 to calculate the pH: Kb=KwKa=1.0\u00d710\u2212147.2\u00d710\u22124\u22481.39\u00d710\u221211K_b = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{7.2 \\times 10^{-4}} \\approx 1.39 \\times 10^{-11}Kb\u200b=Ka\u200bKw\u200b\u200b=7.2\u00d710\u221241.0\u00d710\u221214\u200b\u22481.39\u00d710\u221211<\/p>\n\n\n\n
For the hydrolysis of F\u2212F^-F\u2212: Kb=[OH\u2212]2[F\u2212]K_b = \\frac{[OH^-]^2}{[F^-]}Kb\u200b=[F\u2212][OH\u2212]2\u200b<\/p>\n\n\n\n
Solving for [OH\u2212][OH^-][OH\u2212]: [OH\u2212]=Kb\u00d7[F\u2212][OH^-] = \\sqrt{K_b \\times [F^-]}[OH\u2212]=Kb\u200b\u00d7[F\u2212]\u200b<\/p>\n\n\n\n
Substitute the values: [OH\u2212]=(1.39\u00d710\u221211)\u00d70.301\u22486.77\u00d710\u22126\u2009M[OH^-] = \\sqrt{(1.39 \\times 10^{-11}) \\times 0.301} \\approx 6.77 \\times 10^{-6} \\, \\text{M}[OH\u2212]=(1.39\u00d710\u221211)\u00d70.301\u200b\u22486.77\u00d710\u22126M<\/p>\n\n\n\n
Now, calculate the pOH: pOH=\u2212log\u2061([OH\u2212])=\u2212log\u2061(6.77\u00d710\u22126)\u22485.17\\text{pOH} = -\\log([OH^-]) = -\\log(6.77 \\times 10^{-6}) \\approx 5.17pOH=\u2212log([OH\u2212])=\u2212log(6.77\u00d710\u22126)\u22485.17<\/p>\n\n\n\n
Finally, the pH is: pH=14\u2212pOH=14\u22125.17=8.83\\text{pH} = 14 – \\text{pOH} = 14 – 5.17 = 8.83pH=14\u2212pOH=14\u22125.17=8.83<\/p>\n\n\n\n
\n\n\n\n5. After the addition of 85.4 mL of KOH<\/strong><\/h3>\n\n\n\nThe total volume of KOH added is 85.4 mL.<\/p>\n\n\n\n
\n- Moles of KOH added:<\/li>\n<\/ul>\n\n\n\n
moles of KOH=0.301\u2009M\u00d70.0854\u2009L=0.0257\u2009mol\\text{moles of KOH} = 0.301 \\, \\text{M} \\times 0.0854 \\, \\text{L} = 0.0257 \\, \\text{mol}moles of KOH=0.301M\u00d70.0854L=0.0257mol<\/p>\n\n\n\n
At this point, the amount of KOH added exceeds the moles of HF, so the solution will be basic due to excess OH\u2212OH^-OH\u2212. The excess moles of KOH are: excess moles of KOH=0.0257\u2009mol\u22120.0171\u2009mol=0.0086\u2009mol\\text{excess moles of KOH} = 0.0257 \\, \\text{mol} – 0.0171 \\, \\text{mol} = 0.0086 \\, \\text{mol}excess moles of KOH=0.0257mol\u22120.0171mol=0.0086mol<\/p>\n\n\n\n
The concentration of OH\u2212OH^-OH\u2212 in the total volume is: total volume=56.9\u2009mL+85.4\u2009mL=142.3\u2009mL=0.1423\u2009L\\text{total volume} = 56.9 \\, \\text{mL} + 85.4 \\, \\text{mL} = 142.3 \\, \\text{mL} = 0.1423 \\, \\text{L}total volume=56.9mL+85.4mL=142.3mL=0.1423L [OH\u2212]=0.0086\u2009mol0.1423\u2009L=0.0605\u2009M[OH^-] = \\frac{0.0086 \\, \\text{mol}}{0.1423 \\, \\text{L}} = 0.0605 \\, \\text{M}[OH\u2212]=0.1423L0.0086mol\u200b=0.0605M<\/p>\n\n\n\n
Now, calculate the pOH: pOH=\u2212log\u2061(0.0605)\u22481.22\\text{pOH} = -\\log(0.0605) \\approx 1.22pOH=\u2212log(0.0605)\u22481.22<\/p>\n\n\n\n
Finally, calculate the pH: pH=14\u22121.22=12.78\\text{pH} = 14 – 1.22 = 12.78pH=14\u22121.22=12.78<\/p>\n\n\n\n
\n\n\n\nSummary of pH at each point:<\/h3>\n\n\n\n\n- Before the addition of KOH: pH \u2248 1.83<\/strong><\/li>\n\n\n\n
- After adding 14.0 mL of KOH: pH \u2248 2.65<\/strong><\/li>\n\n\n\n
- At the half-equivalence point: pH = 3.14<\/strong><\/li>\n\n\n\n
- At the equivalence point: pH \u2248 8.83<\/strong><\/li>\n\n\n\n
- After adding 85.4 m<\/li>\n<\/ol>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"Calculate pH for a weak acid\/strong base titration. Determine the pH during the titration of 56.9 mL of 0.301 M hydrofluoric acid (Ka = 7.2*10^4) by 0.301 M KOH at the following points: 1. Before the addition of any KOH 2. After the addition of 14.0 mL of KOH 3. At the half-equivalence point (the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-245411","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245411","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=245411"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245411\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245411"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245411"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245411"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
1. Before the addition of any KOH (Initial pH of HF)<\/strong><\/h3>\n\n\n\nFor the initial pH of a weak acid, we need to use the expression for the dissociation of HF: HF\u21ccH++F\u2212HF \\rightleftharpoons H^+ + F^-HF\u21ccH++F\u2212<\/p>\n\n\n\n
The equilibrium expression for the dissociation is: Ka=[H+][F\u2212][HF]K_a = \\frac{[H^+][F^-]}{[HF]}Ka\u200b=[HF][H+][F\u2212]\u200b<\/p>\n\n\n\n
Let\u2019s assume the concentration of [H+][H^+][H+] is xxx, and at equilibrium: Ka=x2[HF]\u2212xK_a = \\frac{x^2}{[HF] – x}Ka\u200b=[HF]\u2212xx2\u200b<\/p>\n\n\n\n
For weak acids, xxx will be very small compared to the initial concentration, so we can approximate [HF]\u2212x\u2248[HF][HF] – x \\approx [HF][HF]\u2212x\u2248[HF].<\/p>\n\n\n\n
Thus: Ka=x2[HF]K_a = \\frac{x^2}{[HF]}Ka\u200b=[HF]x2\u200b<\/p>\n\n\n\n
Solving for xxx, the concentration of H+H^+H+: x=Ka\u00d7[HF]x = \\sqrt{K_a \\times [HF]}x=Ka\u200b\u00d7[HF]\u200b<\/p>\n\n\n\n
Substitute the values: x=(7.2\u00d710\u22124)\u00d70.301x = \\sqrt{(7.2 \\times 10^{-4}) \\times 0.301}x=(7.2\u00d710\u22124)\u00d70.301\u200b x\u22481.48\u00d710\u22122\u2009Mx \\approx 1.48 \\times 10^{-2} \\, \\text{M}x\u22481.48\u00d710\u22122M<\/p>\n\n\n\n
Now, calculate the pH: pH=\u2212log\u2061([H+])=\u2212log\u2061(1.48\u00d710\u22122)\u22481.83\\text{pH} = -\\log([H^+]) = -\\log(1.48 \\times 10^{-2}) \\approx 1.83pH=\u2212log([H+])=\u2212log(1.48\u00d710\u22122)\u22481.83<\/p>\n\n\n\n
\n\n\n\n2. After the addition of 14.0 mL of KOH<\/strong><\/h3>\n\n\n\nTo calculate the pH after adding 14.0 mL of KOH, we first determine the moles of both HF and KOH.<\/p>\n\n\n\n
\n- Moles of HF:<\/li>\n<\/ul>\n\n\n\n
moles of HF=0.301\u2009M\u00d70.0569\u2009L=0.0171\u2009mol\\text{moles of HF} = 0.301 \\, \\text{M} \\times 0.0569 \\, \\text{L} = 0.0171 \\, \\text{mol}moles of HF=0.301M\u00d70.0569L=0.0171mol<\/p>\n\n\n\n
\n- Moles of KOH:<\/li>\n<\/ul>\n\n\n\n
moles of KOH=0.301\u2009M\u00d70.0140\u2009L=0.00421\u2009mol\\text{moles of KOH} = 0.301 \\, \\text{M} \\times 0.0140 \\, \\text{L} = 0.00421 \\, \\text{mol}moles of KOH=0.301M\u00d70.0140L=0.00421mol<\/p>\n\n\n\n
KOH is a strong base, and it will neutralize HF, forming F\u2212F^-F\u2212 and H2OH_2OH2\u200bO.<\/p>\n\n\n\n
\n- Remaining moles of HF:<\/li>\n<\/ul>\n\n\n\n
0.0171\u2009mol\u22120.00421\u2009mol=0.0129\u2009mol0.0171 \\, \\text{mol} – 0.00421 \\, \\text{mol} = 0.0129 \\, \\text{mol}0.0171mol\u22120.00421mol=0.0129mol<\/p>\n\n\n\n
\n- Moles of F\u2212F^-F\u2212 formed:<\/li>\n<\/ul>\n\n\n\n
0.00421\u2009mol0.00421 \\, \\text{mol}0.00421mol<\/p>\n\n\n\n
Now, calculate the new concentrations in the total volume of the solution, which is: Vtotal=56.9\u2009mL+14.0\u2009mL=70.9\u2009mL=0.0709\u2009LV_{\\text{total}} = 56.9 \\, \\text{mL} + 14.0 \\, \\text{mL} = 70.9 \\, \\text{mL} = 0.0709 \\, \\text{L}Vtotal\u200b=56.9mL+14.0mL=70.9mL=0.0709L<\/p>\n\n\n\n
\n- Concentration of HF:<\/li>\n<\/ul>\n\n\n\n
[HF]=0.0129\u2009mol0.0709\u2009L=0.182\u2009M[HF] = \\frac{0.0129 \\, \\text{mol}}{0.0709 \\, \\text{L}} = 0.182 \\, \\text{M}[HF]=0.0709L0.0129mol\u200b=0.182M<\/p>\n\n\n\n
\n- Concentration of F\u2212F^-F\u2212:<\/li>\n<\/ul>\n\n\n\n
[F\u2212]=0.00421\u2009mol0.0709\u2009L=0.0594\u2009M[F^-] = \\frac{0.00421 \\, \\text{mol}}{0.0709 \\, \\text{L}} = 0.0594 \\, \\text{M}[F\u2212]=0.0709L0.00421mol\u200b=0.0594M<\/p>\n\n\n\n
At this point, we use the Henderson-Hasselbalch equation to find the pH: pH=pKa+log\u2061([F\u2212][HF])\\text{pH} = \\text{pKa} + \\log \\left( \\frac{[F^-]}{[HF]} \\right)pH=pKa+log([HF][F\u2212]\u200b)<\/p>\n\n\n\n
First, calculate pKa\\text{pKa}pKa: pKa=\u2212log\u2061(Ka)=\u2212log\u2061(7.2\u00d710\u22124)\u22483.14\\text{pKa} = -\\log(K_a) = -\\log(7.2 \\times 10^{-4}) \\approx 3.14pKa=\u2212log(Ka\u200b)=\u2212log(7.2\u00d710\u22124)\u22483.14<\/p>\n\n\n\n
Then, calculate the pH: pH=3.14+log\u2061(0.05940.182)\u22483.14+log\u2061(0.326)\u22483.14\u22120.486\u22482.65\\text{pH} = 3.14 + \\log \\left( \\frac{0.0594}{0.182} \\right) \\approx 3.14 + \\log(0.326) \\approx 3.14 – 0.486 \\approx 2.65pH=3.14+log(0.1820.0594\u200b)\u22483.14+log(0.326)\u22483.14\u22120.486\u22482.65<\/p>\n\n\n\n
\n\n\n\n3. At the half-equivalence point (the titration midpoint)<\/strong><\/h3>\n\n\n\nAt the half-equivalence point, the moles of KOH added will be equal to half the moles of HF.<\/p>\n\n\n\n
\n- Moles of KOH at half-equivalence:<\/li>\n<\/ul>\n\n\n\n
0.01712=0.00855\u2009mol\\frac{0.0171}{2} = 0.00855 \\, \\text{mol}20.0171\u200b=0.00855mol<\/p>\n\n\n\n
The concentration of F\u2212F^-F\u2212 and HFHFHF will be the same at this point, so: pH=pKa=3.14\\text{pH} = \\text{pKa} = 3.14pH=pKa=3.14<\/p>\n\n\n\n
\n\n\n\n4. At the equivalence point<\/strong><\/h3>\n\n\n\nAt the equivalence point, all HF has reacted with KOH. The solution now contains only the conjugate base F\u2212F^-F\u2212, which will hydrolyze in water to form OH\u2212OH^-OH\u2212. The hydrolysis reaction is: F\u2212+H2O\u21ccHF+OH\u2212F^- + H_2O \\rightleftharpoons HF + OH^-F\u2212+H2\u200bO\u21ccHF+OH\u2212<\/p>\n\n\n\n
We can use the Kb of F\u2212F^-F\u2212 to calculate the pH: Kb=KwKa=1.0\u00d710\u2212147.2\u00d710\u22124\u22481.39\u00d710\u221211K_b = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{7.2 \\times 10^{-4}} \\approx 1.39 \\times 10^{-11}Kb\u200b=Ka\u200bKw\u200b\u200b=7.2\u00d710\u221241.0\u00d710\u221214\u200b\u22481.39\u00d710\u221211<\/p>\n\n\n\n
For the hydrolysis of F\u2212F^-F\u2212: Kb=[OH\u2212]2[F\u2212]K_b = \\frac{[OH^-]^2}{[F^-]}Kb\u200b=[F\u2212][OH\u2212]2\u200b<\/p>\n\n\n\n
Solving for [OH\u2212][OH^-][OH\u2212]: [OH\u2212]=Kb\u00d7[F\u2212][OH^-] = \\sqrt{K_b \\times [F^-]}[OH\u2212]=Kb\u200b\u00d7[F\u2212]\u200b<\/p>\n\n\n\n
Substitute the values: [OH\u2212]=(1.39\u00d710\u221211)\u00d70.301\u22486.77\u00d710\u22126\u2009M[OH^-] = \\sqrt{(1.39 \\times 10^{-11}) \\times 0.301} \\approx 6.77 \\times 10^{-6} \\, \\text{M}[OH\u2212]=(1.39\u00d710\u221211)\u00d70.301\u200b\u22486.77\u00d710\u22126M<\/p>\n\n\n\n
Now, calculate the pOH: pOH=\u2212log\u2061([OH\u2212])=\u2212log\u2061(6.77\u00d710\u22126)\u22485.17\\text{pOH} = -\\log([OH^-]) = -\\log(6.77 \\times 10^{-6}) \\approx 5.17pOH=\u2212log([OH\u2212])=\u2212log(6.77\u00d710\u22126)\u22485.17<\/p>\n\n\n\n
Finally, the pH is: pH=14\u2212pOH=14\u22125.17=8.83\\text{pH} = 14 – \\text{pOH} = 14 – 5.17 = 8.83pH=14\u2212pOH=14\u22125.17=8.83<\/p>\n\n\n\n
\n\n\n\n5. After the addition of 85.4 mL of KOH<\/strong><\/h3>\n\n\n\nThe total volume of KOH added is 85.4 mL.<\/p>\n\n\n\n
\n- Moles of KOH added:<\/li>\n<\/ul>\n\n\n\n
moles of KOH=0.301\u2009M\u00d70.0854\u2009L=0.0257\u2009mol\\text{moles of KOH} = 0.301 \\, \\text{M} \\times 0.0854 \\, \\text{L} = 0.0257 \\, \\text{mol}moles of KOH=0.301M\u00d70.0854L=0.0257mol<\/p>\n\n\n\n
At this point, the amount of KOH added exceeds the moles of HF, so the solution will be basic due to excess OH\u2212OH^-OH\u2212. The excess moles of KOH are: excess moles of KOH=0.0257\u2009mol\u22120.0171\u2009mol=0.0086\u2009mol\\text{excess moles of KOH} = 0.0257 \\, \\text{mol} – 0.0171 \\, \\text{mol} = 0.0086 \\, \\text{mol}excess moles of KOH=0.0257mol\u22120.0171mol=0.0086mol<\/p>\n\n\n\n
The concentration of OH\u2212OH^-OH\u2212 in the total volume is: total volume=56.9\u2009mL+85.4\u2009mL=142.3\u2009mL=0.1423\u2009L\\text{total volume} = 56.9 \\, \\text{mL} + 85.4 \\, \\text{mL} = 142.3 \\, \\text{mL} = 0.1423 \\, \\text{L}total volume=56.9mL+85.4mL=142.3mL=0.1423L [OH\u2212]=0.0086\u2009mol0.1423\u2009L=0.0605\u2009M[OH^-] = \\frac{0.0086 \\, \\text{mol}}{0.1423 \\, \\text{L}} = 0.0605 \\, \\text{M}[OH\u2212]=0.1423L0.0086mol\u200b=0.0605M<\/p>\n\n\n\n
Now, calculate the pOH: pOH=\u2212log\u2061(0.0605)\u22481.22\\text{pOH} = -\\log(0.0605) \\approx 1.22pOH=\u2212log(0.0605)\u22481.22<\/p>\n\n\n\n
Finally, calculate the pH: pH=14\u22121.22=12.78\\text{pH} = 14 – 1.22 = 12.78pH=14\u22121.22=12.78<\/p>\n\n\n\n
\n\n\n\nSummary of pH at each point:<\/h3>\n\n\n\n\n- Before the addition of KOH: pH \u2248 1.83<\/strong><\/li>\n\n\n\n
- After adding 14.0 mL of KOH: pH \u2248 2.65<\/strong><\/li>\n\n\n\n
- At the half-equivalence point: pH = 3.14<\/strong><\/li>\n\n\n\n
- At the equivalence point: pH \u2248 8.83<\/strong><\/li>\n\n\n\n
- After adding 85.4 m<\/li>\n<\/ol>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"Calculate pH for a weak acid\/strong base titration. Determine the pH during the titration of 56.9 mL of 0.301 M hydrofluoric acid (Ka = 7.2*10^4) by 0.301 M KOH at the following points: 1. Before the addition of any KOH 2. After the addition of 14.0 mL of KOH 3. At the half-equivalence point (the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-245411","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245411","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=245411"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245411\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245411"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245411"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245411"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
To calculate the pH after adding 14.0 mL of KOH, we first determine the moles of both HF and KOH.<\/p>\n\n\n\n
- \n
- Moles of HF:<\/li>\n<\/ul>\n\n\n\n
moles of HF=0.301\u2009M\u00d70.0569\u2009L=0.0171\u2009mol\\text{moles of HF} = 0.301 \\, \\text{M} \\times 0.0569 \\, \\text{L} = 0.0171 \\, \\text{mol}moles of HF=0.301M\u00d70.0569L=0.0171mol<\/p>\n\n\n\n
- \n
- Moles of KOH:<\/li>\n<\/ul>\n\n\n\n
moles of KOH=0.301\u2009M\u00d70.0140\u2009L=0.00421\u2009mol\\text{moles of KOH} = 0.301 \\, \\text{M} \\times 0.0140 \\, \\text{L} = 0.00421 \\, \\text{mol}moles of KOH=0.301M\u00d70.0140L=0.00421mol<\/p>\n\n\n\n
KOH is a strong base, and it will neutralize HF, forming F\u2212F^-F\u2212 and H2OH_2OH2\u200bO.<\/p>\n\n\n\n
- \n
- Remaining moles of HF:<\/li>\n<\/ul>\n\n\n\n
0.0171\u2009mol\u22120.00421\u2009mol=0.0129\u2009mol0.0171 \\, \\text{mol} – 0.00421 \\, \\text{mol} = 0.0129 \\, \\text{mol}0.0171mol\u22120.00421mol=0.0129mol<\/p>\n\n\n\n
- \n
- Moles of F\u2212F^-F\u2212 formed:<\/li>\n<\/ul>\n\n\n\n
0.00421\u2009mol0.00421 \\, \\text{mol}0.00421mol<\/p>\n\n\n\n
Now, calculate the new concentrations in the total volume of the solution, which is: Vtotal=56.9\u2009mL+14.0\u2009mL=70.9\u2009mL=0.0709\u2009LV_{\\text{total}} = 56.9 \\, \\text{mL} + 14.0 \\, \\text{mL} = 70.9 \\, \\text{mL} = 0.0709 \\, \\text{L}Vtotal\u200b=56.9mL+14.0mL=70.9mL=0.0709L<\/p>\n\n\n\n
- \n
- Concentration of HF:<\/li>\n<\/ul>\n\n\n\n
[HF]=0.0129\u2009mol0.0709\u2009L=0.182\u2009M[HF] = \\frac{0.0129 \\, \\text{mol}}{0.0709 \\, \\text{L}} = 0.182 \\, \\text{M}[HF]=0.0709L0.0129mol\u200b=0.182M<\/p>\n\n\n\n
- \n
- Concentration of F\u2212F^-F\u2212:<\/li>\n<\/ul>\n\n\n\n
[F\u2212]=0.00421\u2009mol0.0709\u2009L=0.0594\u2009M[F^-] = \\frac{0.00421 \\, \\text{mol}}{0.0709 \\, \\text{L}} = 0.0594 \\, \\text{M}[F\u2212]=0.0709L0.00421mol\u200b=0.0594M<\/p>\n\n\n\n
At this point, we use the Henderson-Hasselbalch equation to find the pH: pH=pKa+log\u2061([F\u2212][HF])\\text{pH} = \\text{pKa} + \\log \\left( \\frac{[F^-]}{[HF]} \\right)pH=pKa+log([HF][F\u2212]\u200b)<\/p>\n\n\n\n
First, calculate pKa\\text{pKa}pKa: pKa=\u2212log\u2061(Ka)=\u2212log\u2061(7.2\u00d710\u22124)\u22483.14\\text{pKa} = -\\log(K_a) = -\\log(7.2 \\times 10^{-4}) \\approx 3.14pKa=\u2212log(Ka\u200b)=\u2212log(7.2\u00d710\u22124)\u22483.14<\/p>\n\n\n\n
Then, calculate the pH: pH=3.14+log\u2061(0.05940.182)\u22483.14+log\u2061(0.326)\u22483.14\u22120.486\u22482.65\\text{pH} = 3.14 + \\log \\left( \\frac{0.0594}{0.182} \\right) \\approx 3.14 + \\log(0.326) \\approx 3.14 – 0.486 \\approx 2.65pH=3.14+log(0.1820.0594\u200b)\u22483.14+log(0.326)\u22483.14\u22120.486\u22482.65<\/p>\n\n\n\n
\n\n\n\n3. At the half-equivalence point (the titration midpoint)<\/strong><\/h3>\n\n\n\n
At the half-equivalence point, the moles of KOH added will be equal to half the moles of HF.<\/p>\n\n\n\n
- \n
- Moles of KOH at half-equivalence:<\/li>\n<\/ul>\n\n\n\n
0.01712=0.00855\u2009mol\\frac{0.0171}{2} = 0.00855 \\, \\text{mol}20.0171\u200b=0.00855mol<\/p>\n\n\n\n
The concentration of F\u2212F^-F\u2212 and HFHFHF will be the same at this point, so: pH=pKa=3.14\\text{pH} = \\text{pKa} = 3.14pH=pKa=3.14<\/p>\n\n\n\n
\n\n\n\n4. At the equivalence point<\/strong><\/h3>\n\n\n\n
At the equivalence point, all HF has reacted with KOH. The solution now contains only the conjugate base F\u2212F^-F\u2212, which will hydrolyze in water to form OH\u2212OH^-OH\u2212. The hydrolysis reaction is: F\u2212+H2O\u21ccHF+OH\u2212F^- + H_2O \\rightleftharpoons HF + OH^-F\u2212+H2\u200bO\u21ccHF+OH\u2212<\/p>\n\n\n\n
We can use the Kb of F\u2212F^-F\u2212 to calculate the pH: Kb=KwKa=1.0\u00d710\u2212147.2\u00d710\u22124\u22481.39\u00d710\u221211K_b = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{7.2 \\times 10^{-4}} \\approx 1.39 \\times 10^{-11}Kb\u200b=Ka\u200bKw\u200b\u200b=7.2\u00d710\u221241.0\u00d710\u221214\u200b\u22481.39\u00d710\u221211<\/p>\n\n\n\n
For the hydrolysis of F\u2212F^-F\u2212: Kb=[OH\u2212]2[F\u2212]K_b = \\frac{[OH^-]^2}{[F^-]}Kb\u200b=[F\u2212][OH\u2212]2\u200b<\/p>\n\n\n\n
Solving for [OH\u2212][OH^-][OH\u2212]: [OH\u2212]=Kb\u00d7[F\u2212][OH^-] = \\sqrt{K_b \\times [F^-]}[OH\u2212]=Kb\u200b\u00d7[F\u2212]\u200b<\/p>\n\n\n\n
Substitute the values: [OH\u2212]=(1.39\u00d710\u221211)\u00d70.301\u22486.77\u00d710\u22126\u2009M[OH^-] = \\sqrt{(1.39 \\times 10^{-11}) \\times 0.301} \\approx 6.77 \\times 10^{-6} \\, \\text{M}[OH\u2212]=(1.39\u00d710\u221211)\u00d70.301\u200b\u22486.77\u00d710\u22126M<\/p>\n\n\n\n
Now, calculate the pOH: pOH=\u2212log\u2061([OH\u2212])=\u2212log\u2061(6.77\u00d710\u22126)\u22485.17\\text{pOH} = -\\log([OH^-]) = -\\log(6.77 \\times 10^{-6}) \\approx 5.17pOH=\u2212log([OH\u2212])=\u2212log(6.77\u00d710\u22126)\u22485.17<\/p>\n\n\n\n
Finally, the pH is: pH=14\u2212pOH=14\u22125.17=8.83\\text{pH} = 14 – \\text{pOH} = 14 – 5.17 = 8.83pH=14\u2212pOH=14\u22125.17=8.83<\/p>\n\n\n\n
\n\n\n\n5. After the addition of 85.4 mL of KOH<\/strong><\/h3>\n\n\n\n
The total volume of KOH added is 85.4 mL.<\/p>\n\n\n\n
- \n
- Moles of KOH added:<\/li>\n<\/ul>\n\n\n\n
moles of KOH=0.301\u2009M\u00d70.0854\u2009L=0.0257\u2009mol\\text{moles of KOH} = 0.301 \\, \\text{M} \\times 0.0854 \\, \\text{L} = 0.0257 \\, \\text{mol}moles of KOH=0.301M\u00d70.0854L=0.0257mol<\/p>\n\n\n\n
At this point, the amount of KOH added exceeds the moles of HF, so the solution will be basic due to excess OH\u2212OH^-OH\u2212. The excess moles of KOH are: excess moles of KOH=0.0257\u2009mol\u22120.0171\u2009mol=0.0086\u2009mol\\text{excess moles of KOH} = 0.0257 \\, \\text{mol} – 0.0171 \\, \\text{mol} = 0.0086 \\, \\text{mol}excess moles of KOH=0.0257mol\u22120.0171mol=0.0086mol<\/p>\n\n\n\n
The concentration of OH\u2212OH^-OH\u2212 in the total volume is: total volume=56.9\u2009mL+85.4\u2009mL=142.3\u2009mL=0.1423\u2009L\\text{total volume} = 56.9 \\, \\text{mL} + 85.4 \\, \\text{mL} = 142.3 \\, \\text{mL} = 0.1423 \\, \\text{L}total volume=56.9mL+85.4mL=142.3mL=0.1423L [OH\u2212]=0.0086\u2009mol0.1423\u2009L=0.0605\u2009M[OH^-] = \\frac{0.0086 \\, \\text{mol}}{0.1423 \\, \\text{L}} = 0.0605 \\, \\text{M}[OH\u2212]=0.1423L0.0086mol\u200b=0.0605M<\/p>\n\n\n\n
Now, calculate the pOH: pOH=\u2212log\u2061(0.0605)\u22481.22\\text{pOH} = -\\log(0.0605) \\approx 1.22pOH=\u2212log(0.0605)\u22481.22<\/p>\n\n\n\n
Finally, calculate the pH: pH=14\u22121.22=12.78\\text{pH} = 14 – 1.22 = 12.78pH=14\u22121.22=12.78<\/p>\n\n\n\n
\n\n\n\nSummary of pH at each point:<\/h3>\n\n\n\n
- \n
- Before the addition of KOH: pH \u2248 1.83<\/strong><\/li>\n\n\n\n
- After adding 14.0 mL of KOH: pH \u2248 2.65<\/strong><\/li>\n\n\n\n
- At the half-equivalence point: pH = 3.14<\/strong><\/li>\n\n\n\n
- At the equivalence point: pH \u2248 8.83<\/strong><\/li>\n\n\n\n
- After adding 85.4 m<\/li>\n<\/ol>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"Calculate pH for a weak acid\/strong base titration. Determine the pH during the titration of 56.9 mL of 0.301 M hydrofluoric acid (Ka = 7.2*10^4) by 0.301 M KOH at the following points: 1. Before the addition of any KOH 2. After the addition of 14.0 mL of KOH 3. At the half-equivalence point (the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-245411","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245411","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=245411"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245411\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245411"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245411"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245411"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- After adding 14.0 mL of KOH: pH \u2248 2.65<\/strong><\/li>\n\n\n\n
- Before the addition of KOH: pH \u2248 1.83<\/strong><\/li>\n\n\n\n
- Moles of KOH added:<\/li>\n<\/ul>\n\n\n\n
- Moles of KOH at half-equivalence:<\/li>\n<\/ul>\n\n\n\n
- Concentration of F\u2212F^-F\u2212:<\/li>\n<\/ul>\n\n\n\n
- Concentration of HF:<\/li>\n<\/ul>\n\n\n\n
- Moles of F\u2212F^-F\u2212 formed:<\/li>\n<\/ul>\n\n\n\n
- Remaining moles of HF:<\/li>\n<\/ul>\n\n\n\n
- Moles of KOH:<\/li>\n<\/ul>\n\n\n\n